At time `t=0` a body is thrown vertically upwards with a velocity "50m/sec" .At time t=2 sec another body is thrown vertically upwards with the same velocity "50m/sec" .the two bodies will meet at time__ sec `(a=10m/sec^(2))`

Matching type Question : For a body thrown vertically upwards with initial velocity u

A body A is projected upwards with a velocity of 90m//s . The second body B is projected upwards with the same initial velocity but after 4 sec. Both the bodies will meet after

A body is projected vertically upwards wth a velocity u=5m//s . After time t another body is projected vertically upward from the same point with a velocity v=3m//s . If they meet in minimum time duration measured from the projection of first body, then t=k/g sec find k (where g is gravitatio acceleration).

A body is thrown vertically upwards which graph represents variation of velocity w.r.t time

A body is thrown vertically upward with a velocity of 9.8 m/s. Calculate the maximum height attained by the body. (g=9.8m//s^(2))

Velocity-time graph of a body thrown vertically up is

Velocity-time curve for a body projected vertically upwards is

A body is thrown vertically upwards and rises to a height of 10m. The velocity with which the body was thrown is (g - 9.8m/( s^(2) )

Calculate the maximum height attained by the body thrown vertically upward with a velocity of 9.8m//s (g=9.8m//s^2)